Question

14–2. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 14.1. 100 65 MPa mod sd niog 1622 B Prob. 14-2

Use protractor, compass, draw the MOHR'S CIRCLE. For scale, use millimeter. For example 65 MPa = 6.5 centimeter, and 20 MPa = 2 centimeter.

Please write in order and write neat.

Answer 1

Points (or you to ( 65,- 203 State Ceutore OutOy = 6570 - 3215mpo impa=imim Radius - Jay-c10 2+ Cry + Zuy 2 - 1 6524402 = 38116 MPa z (mpa) 50 Zmau = 38.16 MPa tot 133,571 mpa (0,20) 1/ 148.39° 110 lo Klod (325,0) Top (70.66) JB (-5.66)11 10 20 30 40 50 6065 o(mpa 100 l mp4 V 120 W (65,-20) on=31:429 mpa angleo making from verticay cruis = 90-30=600 in mong circle it makes = 20=120° at 0=60° Th=311429 MPa = 331571 Mas Int=38.146 MPa ccoon n face the

additional information by use of software

Mohr's Circle Transformations Module Back File Analysis Options Failure Theories Help x direction y direction Normal Stress Normal Stress 0.000 65.000 Principal Stress Orientation Max In-plane Shear Stress Orientation -xy direction Shear Stress 20.000 Tension (+) Compression (-) Tension (+) Compression - CW on x face C CCW on x face (+) Compute Details MPa 29.200 Show Absolute Maximum Shear Stress 74.200 Txy Max = 38.161 MPa Txy Mix -148.390 Compute Normal and Shear Stresses at Arbitrary Orientation 0 80 Angle from x axis to n axis 70 60 |60.0 ; Null Compute QOD 120.000 On = 31.429 MPa Tension (+) Ot = 33.571 MPa Tension (+) Tnt = 38.146 MPa CCW on n face (+) Opi = 70.661 MPa Op2 = -5.661 MPa C = 32.500 MPa R = 38.161 MPa -90 -80 -70 Type here to search 2 4 » - 08:33 AM GENG 22-04-2020 w

Mohr's Circle Stress On Unrotated Element 60.00 Normal Stress, X: 50.00 (33.57, 38.15) Max shear = 38.16 40.00 + Normal Stress, Y: 65 30.00 (0.000,20.00) Shear Stress on Y Face: -20 20.00 Rotation Angle 10.00 Min prin=-5.661 Specify the angle of the rotated element. Angles are with respect to the horizontal axis, and positive angles are counterclockwise. -40.0 (32.50, 0.0) 40.0 0 -20. 0 0 20.0 60.0 80.0 100.0 Max prin = 70.66 -10.00+ Tip: You can mouse drag OR use keyboard arrow keys to adjust angle. -90 1 90 -20.00 Rotation Angle: 60 deg -30.00 (65.00, -20.00) Results -40.00 + -50.00+ Min shear = -38.16 (31.43, -38.15) Angle (deg] 60.00 150.0 -60.001 Original Stress - Principal Stress - Transformed Stress - Extreme Shear Stress Rotated Element, Normal Stress X Rotated Element, Normal Stress Y Rotated Element, Shear Stress Max Principal Stress Min Principal Stress Max Extreme Shear Stress Min Extreme Shear Stress Value 31.43 33.57 38.15 70.66 -5.661 38.16 -38.16 - x: 93.60, y: 39.23 -74.20 15.80 -29.20 60.80

Unrotated Element(Black Line in Mohr's Circle) Rotated Element (60.00 deg) 0.X = 31.43 o.y = 65.00 T= -38.15 T= -20.00 o.y = 33.57 T=38/15 T = 20.000.000 16x=0l doo IT= 20.00 T = 38.15 T= -20.00 o.y = 33.57 1-38.15 0.X = 31.43 o.y = 65.00 Element Rotated to Max Principal Stress (-74.20 deg) Element Rotated to Max Shear Stress (-29.20 deg) o.y = 32.50 0.x = 70.66 0.3 33.98 T= -38.16 T = 0.000 O.y = 15661 = 0.000 o.y = -5.661 16.000 T = 0.000 L10 0.X = 32.50 T= 38.16 0.X = 70.66 O.y = 32.50

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